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Quindecagon please help

  1. Aug 11, 2004 #1
    How do i show that a group of symmetries of a regular quindecagon has order |G| = 30 ?
    And how do i describe the elements of G and classify them by their order? Thanks
  2. jcsd
  3. Aug 11, 2004 #2

    matt grime

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    Do you know about dihedral groups? In general the symmetries of a regular n-gon are called the dihedral group of order 2n, denoted either [tex]D_n[/tex] or [tex]D_{2n}[/tex].

    There are several ways to think about symmetries of solid figures. Any symmetry must send a vertex (corner) to another corner, and it can map the neighbouring corners in one of two ways. From that you can see it must have order 30 (15 possible points to send it to, and 2 possible orientations). However that doesn't help you describe the elements of the group and find their order.
    Instead, try thinking of some basic operations that generate all the symmetries. In particular pick a line through two of the vertices, let s be reflection in that line, let t be rotation by 360/15 degrees. What are the orders of s and t? what is tst? are all symmetries generated by s and t?
    how many rotations are there, reflections in other axes, have you now counted all of the symmetries, etc.
  4. Sep 17, 2004 #3


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    as Matt said, in general for any figure, to count its symmetries, pick some object on it like a vertex, and ask two questions: 1) how many maps leave that object fixed? 2) how many different such objects can I map that object to?

    then the order of the group is the product of these two numbers. This is a nice example of lagrange's theorem, and a beautiful illustration of the concept of a coset discussed elsewhere in this forum. i.e. the maps leaving a vertex fixed form a subgroup, the "isotropy" group of that object, and the maps taking a given vertex to another specific one form a coset of that subgroup.

    It also illustrates the concept of "conjugation" (passing from x to yxyinverse), since if there is a map sending one vertex to another, then the two subgroups fixing those vertices are conjugate to each other. In particular neither one is "normal" (invariant under conjugation).

    for instance on an icosahedron, there are 20 faces and 12 vertices. if we pick vertices as our objects, then each vertex is common to 5 triangles, so is left fixed by the 5 rotations about that vertex which permute the 5 neighboring triangles cyclically.

    Since there are 12 vertices and any one can be rotated into all 12 of them, the order of the group is (5)(12) = 60.

    as he said as well, the structure of the group is much harder than just knowing the order.

    In this example however, note that all rotations about a vertex have order 5, and there are 12 vertices, one at each ends of the axis through one. Hence there are 24 non trivial rotations of order 5, and one trivial rotation of order 1.

    Now if we consider faces, these are triangles and there are 20 of them also opposite in pairs and sharing the same rotations in pairs, so there are 20 non trivial rotations of order 3.

    Then there are 30 edges, also opposite in pairs giving us 15 rotations of order 2, giving us 20 + 24 + 15 + 1 = 60 elements in all, of orders 1,2,3, and 5. Now by basic theorems, since the factors of the group's order are 2,3, and 5, we do not expect any other elements of any prime order, and we have found 60 elements anyway, so we are done. of course in a more general group there might be elements of non prime order.

    notice as well by our previous remarks that all isotropy groups of the same order are conjugate to each other, hence if a subgroup is invariant under "conjugation" (operation of form x goes to yxyinverse), then it either contains all elements of a given order or none.

    Since also the order of a subgroup divides the order of the group, the only possible orders for such an invariant subgroup would be sums of 1 and some of the numbers 20, 24, 15, and also dividing 60. Since the only such numbers are 1 and 60, this interesting group has no invariant subgroups. It is called "simple" for this reason and is the smallest non abelian simple group.

    the next such has order 168 and is the group of automorphisms of a famous plane curve of degree 4. in a famous book classifying simple groups of small order, this second one was overlooked by C. Jordan, and discovered by F. Klein I believe. It is also a famous finite matrix group, as also noted by Klein.
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